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Chunky
2010-Jul-18, 08:04 AM
I have a question. when you shake a carbonated beverage the pressure in the bottle rises. If you were to open it right then. The excess pressure would spew out the drink, right? (i guess its from the pressure. the instant release of carbon from the drink?) but if you wait a while and let it sit, it will not spew out when you open it. my question is, is the pressure in the shaken up carbonated beverage reduced through the time it sits before you open it?

:D

just wondering..

caveman1917
2010-Jul-18, 08:14 AM
The carbon in the air-part and in the liquid part are in equilibrium. When you shake the liquid, you increase the pressure on the carbon in the liquid part, therefore it will restore equilibrium by increasing pressure in the air part. So now you have high pressure. When you let the liquid settle down again, the pressure on the carbon in the liquid part will go down, and it will start restoring equilibrium once again, back to your starting point.

Chunky
2010-Jul-18, 08:36 AM
The carbon in the air-part and in the liquid part are in equilibrium. When you shake the liquid, you increase the pressure on the carbon in the liquid part, therefore it will restore equilibrium by increasing pressure in the air part. So now you have high pressure. When you let the liquid settle down again, the pressure on the carbon in the liquid part will go down, and it will start restoring equilibrium once again, back to your starting point.

if it is in equilibrium right after it is shaken up, why must it balance back to a( for lack of a better word) neutral state?

isn't the carbon fusing itself back into the beverage?
like... cold fusion?

:D

caveman1917
2010-Jul-18, 08:47 AM
if it is in equilibrium right after it is shaken up, why must it balance back to a( for lack of a better word) neutral state?

isn't the carbon fusing itself back into the beverage?
like... cold fusion?

:D

It's always in equilibrium (technically it will always tend to one). The pressure in the air part will simply follow the pressure in the liquid part. By shaking you increase pressure in the liquid part, and the air part will follow. By letting the liquid settle down you decrease pressure in the liquid part, and again the air part will follow.

There is no 'neutral' state, any state is as good as any other. It's simply that the air will follow the liquid and vice versa. Manipulate one, and you manipulate the other.

Jeff Root
2010-Jul-18, 09:20 AM
A given number of carbon dioxide molecules in the form of
gas, at a given temperature and pressure, has a larger volume
than the same number of carbon dioxide molecules dissolved
in water. So something like 1 liter of carbon dioxide gas can
be dissolved in 1-3/4 liter of liquid water to give 2 liters of
carbonated water. The volume of the solution is much less
than the sum of the volumes of the separate gas and liquid.

That is simply because the molecules of a gas are flying
around, bumping into each other and flying apart again, with
large empty spaces between them, while the molecules of a
liquid (or a solution) are touching each other all the time, like
a huge number of ping pong balls stored in a big box.

One question that really begs to be answered: Why does the
pressure go up when you shake the bottle? I would guess that
carbon dioxide which is dissolved in the water comes out of
solution for some reason, and joins the big bubble at the top
of the bottle. Since carbon dioxide gas is less dense than the
carbon dioxide in solution in water, it "tries" to expand to fill
a larger volume. But since the volume is limited by the bottle,
the pressure in the gas increases, instead. The increased
pressure of the gas causes the amount of carbon dioxide that
was shaken out of solution to be slowly forced back into the
water and back into solution after the shaking stops.

-- Jeff, in Minneapolis

caveman1917
2010-Jul-18, 09:33 AM
One question that really begs to be answered: Why does the
pressure go up when you shake the bottle? I would guess that
carbon dioxide which is dissolved in the water comes out of
solution for some reason, and joins the big bubble at the top
of the bottle. Since carbon dioxide gas is less dense than the
carbon dioxide in solution in water, it "tries" to expand to fill
a larger volume. But since the volume is limited by the bottle,
the pressure in the gas increases, instead. The increased
pressure of the gas causes the amount of carbon dioxide that
was shaken out of solution to be slowly forced back into the
water and back into solution after the shaking stops.

I suppose shaking the liquid will give a high average movement to its constituent molecules, which will thus apply more pressure against the carbon molecules. Then pressure equalizes by carbon escaping into the air part. If the liquid is allowed to settle down again (decrease the average movement of its molecules), the pressure drops, and it is equalized again by carbon going from the air to the liquid. I don't think the relative density of the carbon in both parts matters, it will merely always try to equalize pressure.

Suppose pressure in the air part is bigger than in the liquid part, then there will be a downwards pressure gradient on the molecules (from air to liquid), pushing the molecules from the air to the liquid. And the other way around in the other case.

The main thing is equilibrium of pressure.

If you leave the bottle open, all carbon will dissipate since it will constantly be trying to equalize pressure, but it will always fail. It's not that there is a fixed amount of carbon that tries to come out of the liquid when you shake it, which then "expands" and due to volume increases in pressure, pressure gradients are the driving force, as long as that isn't equalized the system keeps working.

This seems to make the most sense to me, but i'm not an expert either.

Tobin Dax
2010-Jul-18, 10:03 AM
I suppose shaking the liquid will give a high average movement to its constituent molecules, which will thus apply more pressure against the carbon molecules. Then pressure equalizes by carbon escaping into the air part. If the liquid is allowed to settle down again (decrease the average movement of its molecules), the pressure drops, and it is equalized again by carbon going from the air to the liquid. I don't think the relative density of the carbon in both parts matters, it will merely always try to equalize pressure.

After reading that a couple times, I can buy that as an explanation. Bubbles created by shaking the liquid could have lower pressure than the surrounding liquid, causing the release of carbon dioxide into the bubbles. However, the relative density of carbon dioxide in both parts does matter, as it affects the pressure in the gas if not the liquid. If it did not matter, then the pressure of the gas would not decrease as the carbon dioxide went back into solution after the shaking stops.

caveman1917
2010-Jul-18, 10:05 AM
After reading that a couple times, I can buy that as an explanation. Bubbles created by shaking the liquid could have lower pressure than the surrounding liquid, causing the release of carbon dioxide into the bubbles. However, the relative density of carbon dioxide in both parts does matter, as it affects the pressure in the gas if not the liquid. If it did not matter, then the pressure of the gas would not decrease as the carbon dioxide went back into solution after the shaking stops.

Yes relative density matters, but only in so far as it has an effect on pressure, it is secondary not primary.

ETA: it's not really secondary either, it doesnt change anything by itself, it's just one of the ways to effect pressure. For example if we increase the temperature of the gas, that will cause carbon to move into the liquid, if we increase the volume of the gas, that will cause carbon to move out of the liquid, etc

JohnD
2010-Jul-18, 10:34 AM
It's the bubbles, as foci for gas to come out of solution, but any small particles will do.

Open a bottle of something fizzy, carefully. Minor fizz. Even pour it into a glass, still minor fizz, mostly from dust in the glass or imperfetcions on the wall.

Now add something that will increase the number of foci in the drink to form bubbles. Salt, sugar, Mintoes (whatever they are), sand even, or put your hand over the glass and shake it, to increase the air bubbles in the drink. Careful! You will get flooded with fizz!

If you can do that with the drink still in the bottle, and get the cap back on, the the explosion of drink won't happen. Solid particles will sink to the bottom, soluble ones will dissolve and you will be able to open the bottle again later, without the fountain of fizz.

John

caveman1917
2010-Jul-18, 10:42 AM
Interesting, so that's why it takes so long for the carbon to disappear when it's poured in a glass under atmospheric conditions. While it tries to go to equilibrium, the rate at which it does so is hampered by available 'transport mechanisms'.

Tobin Dax
2010-Jul-18, 11:05 AM
It's the bubbles, as foci for gas to come out of solution, but any small particles will do.

That's pretty much what I thought, and what I was hinting at in my post.

Strange
2010-Jul-18, 12:56 PM
One question that really begs to be answered: Why does the
pressure go up when you shake the bottle? I would guess that
carbon dioxide which is dissolved in the water comes out of
solution for some reason, and joins the big bubble at the top
of the bottle. Since carbon dioxide gas is less dense than the
carbon dioxide in solution in water, it "tries" to expand to fill
a larger volume. But since the volume is limited by the bottle,
the pressure in the gas increases, instead. The increased
pressure of the gas causes the amount of carbon dioxide that
was shaken out of solution to be slowly forced back into the
water and back into solution after the shaking stops.

That is pretty much the answer I was going to give.

It's the bubbles, as foci for gas to come out of solution, but any small particles will do.

I'm not convinced. If the bottle is under pressure so that eventually the CO2 released by shaking redissolves, why would it come out of solution even with bubbles (or anything else) as foci? It must be in a stable equilibrium (the amount of CO2 dissolved in the water being dependent on temperature and pressure). caveman's explanation about shaking lowering pressure locally is slightly more plausible. But not much...

Tobin Dax
2010-Jul-18, 03:31 PM
I'm not convinced. If the bottle is under pressure so that eventually the CO2 released by shaking redissolves, why would it come out of solution even with bubbles (or anything else) as foci? It must be in a stable equilibrium (the amount of CO2 dissolved in the water being dependent on temperature and pressure). caveman's explanation about shaking lowering pressure locally is slightly more plausible. But not much...

The short version is that shaking the bottle ruins that equilibrium, which allows the CO2 to be released from the liquid.
Since shaking is an external phenomenon, it can change the internal state of the bottle and it does. Just after the shaking is over, the gases and liquid in the bottle aren't in a stable equilibrium, but they will reach a stable equilibrium again after some time.

When you shake the bottle, you mix the liquid and air, causing bubbles to form within the liquid (from small regions of newly-trapped air). Bubbles in soda, beer, etc., always act as foci and allow the the CO2 molecules to leave the solution. After the shaking stops and bubbles stop forming, the gaseous CO2 will slowly redissolve into the liquid.

mike alexander
2010-Jul-19, 03:34 AM
The pressure does not go up when you shake the bottle. The pressure of the gas over the liquid in a closed system remains constant (Henry's Law) at a constant temperature.

If you open a bottle of soda that has been allowed to stand the equilibrium is obviously disturbed. After the initial pressure is released the carbon dioxide in solution will eventually escape from the surface of the liquid; since this surface is small, and the carbon dioxide must migrate from deeper in the can the time for escape can be quite long.

When the bottle is shaken the bubbles produced do indeed provide nucleation sites and increased surface area) for growth, but this is more a kinetic effect than a thermodynamic one.

Something similar occurs in supercooled liquids. Extremely pure undisturbed water can be cooled far below its freezing point and remain liquid if no nucleation sites are present; just rapping the side of the container can cause it to just about instantaneously freeze, as shown here (http://il.youtube.com/watch?v=LZ6qRT5bLtI&feature=related).

Antice
2010-Jul-19, 05:17 AM
The same can be done with heating. without any nucleation sites water will not boil and vapor will only form at the surface of the hot water. this also limits the energy transfer and allows water to be heated to well above boiling.

Easy DIY experiment. heat a cup of water to boiling in the microwave oven. let it cool off then heat it again. it will no not boil if you get the conditions right. because all the microscopic air bubbles that tend to be stuck to the container wall has been removed. A word of warning. DO NOT touch the cup with superheated water. it will explosively form steam when disturbed. potentially causing severe injuries.

dgavin
2010-Jul-19, 06:56 PM
Be even more carfull with distilled or fairly pure water in a microwave oven. This was done on Mythbusters once and they managed to cause an explosive flash boil of water when they added an impurity (sugar cube) to it after heating. After the flash boil there was only about 1/5th of the water left in the container.

It was a fairly impressive thing to watch, but i wouldn't do this myself at home. They used a pyrex container that survived the flash boil. A regular drinking glass would of been consumed as part of the explosion, adding rather sharp shrapnel to it.

caveman1917
2010-Jul-19, 11:04 PM
The pressure does not go up when you shake the bottle. The pressure of the gas over the liquid in a closed system remains constant (Henry's Law) at a constant temperature.

Could you elaborate on this?

It would seem to me that if I open a bottle which has just been shaken, a lot more pressure is released than if I open one which has been left undisturbed.
Even when waiting a little for the bubbles to disappear.

ETA: Henry's law would apply to closed systems, but can we consider a bottle to which external forces are applied a closed system?

cjameshuff
2010-Jul-19, 11:20 PM
I don't buy agitating molecules or nucleation sites as explanations. Solubility of gases is temperature dependent, but the temperature change is tiny...and while nucleation sites explain the explosive foaming after the container is opened, they only matter once pressure is released, equilibrium pressure will be independent of surface area (probably some minor surface energy effects, but I doubt they're significant). And the pressure increase is certainly real, you can easily feel the change with a plastic or metal container.

What I suspect happens is that shaking causes local areas of low pressure and high surface area that outgas. This leaves the gas at a higher than equilibrium pressure, but the liquid quickly settles into a state with low surface area and the small CO2-depleted volumes quickly mix into the surrounding liquid, limiting the rate at which the system can return to equilibrium.

caveman1917
2010-Jul-20, 12:17 AM
What I suspect happens is that shaking causes local areas of low pressure and high surface area that outgas. This leaves the gas at a higher than equilibrium pressure, but the liquid quickly settles into a state with low surface area and the small CO2-depleted volumes quickly mix into the surrounding liquid, limiting the rate at which the system can return to equilibrium.

But then why does the liquid also 'squirt' out of the bottle when one opens it? Shouldn't it just be the gas that gets released then, if it is due to the state of non-equilibrium between gas and liquid.

There must be something different about the liquid too, in the post-shake state.

dgavin
2010-Jul-20, 12:58 AM
Think of it in another way, the gas is in a state of supersaturation in the liquid. Disturbances of the fluid, or sonic waves even, will cause a decrease of the saturation/temperature gradients just like they cause a decrease in the freezing points of a fluid. You also have a case where technically the gas isn't really dissolved in water, but is actually suspended between a number of water molecules. (similar to how some minerals seem to dissolve into water, but are actually suspended). It's far far simpler to disturb molecules out of a suspension then out of a dissolved bonding, as the suspension there is not a ionic bond. So the motion induced in movement of the molecules such as shaking, actual causes the suspended gas to be released from the molecules of water.

It's still mostly in the fluid, but no longer in a state of suspension. Some is released into the top of the bottle until the pressure there prevents the un-suspended gas from leaving the water.

At this point opening the bottle causes a sudden decrease in pressure, allowing the still un-suspended gas to vacate the water rapidly, which we see as intense foaming at the top of the solution.

If you let the bottle sit a while though, the un-suspended gasses will eventually be re-suspended, at the same time allowing some of the gas in the top of the water to migrate back into the fluid, until the point of supersaturation for temperature is reached again. At this point when you open the bottle there is only a foaming action from the disturbance of the fluid from the act of opening the bottle. If you could actually open a soda bottle without disturbing the fluid, and without any sonic noise from it, you would see almost no foam on the surface of the fluid right after opening it.

This is also why freezing a soda will cause a can or bottle to explode at times. As the liquid turns solid, almost all the gas is released from suspension causing the can/bottle to to be superpresurized. Can's will almost always explode when frozen, where as bottles rarely.

There are some fluids that actually do dissolve gasses, but water isn't one of them. Any and all gasses in water are suspended, including O2 and C02.

caveman1917
2010-Jul-20, 01:13 AM
[ETA: Nevermind this post, i got two people mixed up :)]

That would clear my objection, the difference in the liquid would be the state of (non)suspension of the carbon within.

So the motion induced in movement of the molecules such as shaking, actual causes the suspended gas to be released from the molecules of water.

I don't buy agitating molecules {...} as explanations.

I'm a bit confused here.

Your idea is testable in a simple way, since it seems to be based on the 'smallness' of the surface area of the liquid. If you shake two bottles, but then leave one in a horizontal position and one in a vertical position, there should be a difference when you open them a little while later. One would have a higher surface area than the other.

cjameshuff
2010-Jul-20, 01:46 AM
But then why does the liquid also 'squirt' out of the bottle when one opens it? Shouldn't it just be the gas that gets released then, if it is due to the state of non-equilibrium between gas and liquid.

There must be something different about the liquid too, in the post-shake state.

As I said, the nucleation sites explain that, they just don't explain the increase of pressure.

caveman1917
2010-Jul-20, 01:53 AM
As I said, the nucleation sites explain that, they just don't explain the increase of pressure.

But it still does this even if we allow a little time for the bubbles to have disappeared.
Or do you mean the nucleation sites in the sense of the un-suspended gas in the liquid that dgavin was referring to?

cjameshuff
2010-Jul-20, 02:15 AM
But it still does this even if we allow a little time for the bubbles to have disappeared.

The CO2 portion of the bubbles has dissolved. Any fraction of N2 and O2 that remains would dissolve more slowly, but produce very small bubbles that would remain in suspension for a considerable amount of time, drag and other forces being much stronger for them than buoyancy forces. While small, they would still provide nucleation sites. When pressure drops due to the container being opened, they expand a bit, serve as nucleation sites for outgassing CO2, and rapidly escape suspension as a fountain of foam. If the container is left to sit for a long time, they dissolve completely or rise to the surface.

Or do you mean the nucleation sites in the sense of the un-suspended gas in the liquid that dgavin was referring to?

dgavin's comments about suspension versus solution don't sound like any description of either I've ever heard. Oxygen, CO2, etc in water behave just like any other gas dissolved in a liquid, with solubility described by Henry's law. A suspension wouldn't have a pressure- and temperature-dependent equilibrium with un-suspended gas, there would be no supersaturation state, and in general dgavin's explanation just doesn't seem to hold together.

caveman1917
2010-Jul-20, 02:24 AM
The CO2 portion of the bubbles has dissolved. Any fraction of N2 and O2 that remains would dissolve more slowly, but produce very small bubbles that would remain in suspension for a considerable amount of time, drag and other forces being much stronger for them than buoyancy forces. While small, they would still provide nucleation sites. When pressure drops due to the container being opened, they expand a bit, serve as nucleation sites for outgassing CO2, and rapidly escape suspension as a fountain of foam. If the container is left to sit for a long time, they dissolve completely or rise to the surface.

So if we have a container where the gas part is pure CO2, it wouldn't show that behaviour?

cjameshuff
2010-Jul-20, 02:35 AM
So if we have a container where the gas part is pure CO2, it wouldn't show that behaviour?

If I'm right, yes. You'd have to degass the water first, and make sure there's no contamination with air.

Something that might throw that prediction off...the surface thermodynamics of those tiny bubbles. Surface tension effects might affect the solubility of the gas trapped in the bubbles, and keep small CO2 bubbles around longer than they'd otherwise last. I'm not sure of the sign of the effect, let alone the magnitude...it may instead make them dissolve even faster than you'd expect from their tiny volume to surface area ratio, or be unnoticeable.

caveman1917
2010-Jul-20, 02:40 AM
If I'm right, yes. You'd have to degass the water first, and make sure there's no contamination with air.

Something that might throw that prediction off...the surface thermodynamics of those tiny bubbles. Surface tension effects might affect the solubility of the gas trapped in the bubbles, and keep small CO2 bubbles around longer than they'd otherwise last. I'm not sure of the sign of the effect, let alone the magnitude...it may instead make them dissolve even faster than you'd expect from their tiny volume to surface area ratio, or be unnoticeable.

Well the surface tension effect should be about the same in both cases (pure CO2 or mixed), so there should at least be a noticeable difference between the two (if not taking away the behaviour altogether).

The general idea could indeed be tested by two bottles, one horizontal and one vertical, though.

dgavin
2010-Jul-20, 07:54 PM
dgavin's comments about suspension versus solution don't sound like any description of either I've ever heard. Oxygen, CO2, etc in water behave just like any other gas dissolved in a liquid, with solubility described by Henry's law. A suspension wouldn't have a pressure- and temperature-dependent equilibrium with un-suspended gas, there would be no supersaturation state, and in general dgavin's explanation just doesn't seem to hold together.

Unless chemestry has changed since my H.S. chem courses, then even O2 and CO2 don't /actualy/ disolve in water, the gas is diffused(?) so much via suspension(?) in water that the gas behaves as if it was disolved.

I think my however the use of the word suspension should actualy be replaced by diffusion in my explanation. I can't remember exactly what the term in chemistry was called but I definately remember that a unique proper of water as a liquid is that while gasses /act/ disolved in it, they actually are not disolved, but diffused(? if i got the right term for it now).

Jeff Root
2010-Jul-20, 08:37 PM
Suspension and diffusion are not only different things, but different
kinds of things. Suspension is a state of being, while diffusion is a
process.

The carbon dioxide diffuses throughout the water and dissolves in it,
resulting in "carbonated water", a solution of water and carbonic acid:

http://en.wikipedia.org/wiki/Carbonated_water
http://en.wikipedia.org/wiki/Solvation

-- Jeff, in Minneapolis

Jeff Root
2010-Jul-20, 08:53 PM
Mike,

I have to think the pressure goes up when you shake the bottle.

You have a bottle mostly filled with carbonated water, with a large
bubble of carbon dioxide, all under a very slight pressure relative to
normal atmospheric pressure. When you shake the bottle, some
of the liquid becomes gas. The gas normally has a much larger
volume than the liquid.

Replace the bottle cap with a rubber balloon. When you shake the
bottle, the gas expands into the balloon, while the volume of the
liquid hardly reduces at all. With the bottle cap on, the gas can't
expand, so the pressure inside the bottle rises instead.

-- Jeff, in Minneapolis

Ara Pacis
2010-Jul-21, 06:22 AM
Does the shaking induce cavitation and does that have any effect on pressure and foaming?

I recall that as a child we would prank each other with shaken bottles and cans of pop. We learned that tapping on the vessel seemed to reduce the foaming. I suspect that this caused side-clinging bubbles to rise to the surface and reduced the deeper bubble formation sites that would cause foaming once pressure was released.

mike alexander
2010-Jul-23, 03:47 AM
Ta-daa! (http://www.wonderquest.com/spewing-soda-can.htm)

A pretty good explanation here! Lots of contributors above have a good piece of the answer, and this pulls it all together. Especially the influence of bubbles.

An important point: For a closed system in dynamic equilibrium like CO2 gas dissolved in water, shaking would not affect the equilibrium. If you accept Henry's Law. The impression that the pressure is greater after shaking is very powerful, but misleading.

cjl
2010-Jul-23, 03:50 AM
Unless chemestry has changed since my H.S. chem courses, then even O2 and CO2 don't /actualy/ disolve in water, the gas is diffused(?) so much via suspension(?) in water that the gas behaves as if it was disolved.

I think my however the use of the word suspension should actualy be replaced by diffusion in my explanation. I can't remember exactly what the term in chemistry was called but I definately remember that a unique proper of water as a liquid is that while gasses /act/ disolved in it, they actually are not disolved, but diffused(? if i got the right term for it now).

CO2 in water definitely dissolves to form carbonic acid - it's a significant part of the flavor of many sodas.

mike alexander
2010-Jul-23, 05:23 AM
Strictly speaking, carbon dioxide gas dissolves in water, which then forms a small amount of carbonic acid in equilibrium with the gas. It's on the order of 1% or so, but that's enough to significantly increase the CO2 water solubility.

But, yes, CO2 definitely forms solvated molecules in water, and carbonic acid, too.

JohnD
2010-Jul-23, 09:21 AM
It may be only a small amount of CO2 that dissolves in water, but it has enormous importance in human and mammalian physiology.
I can't speak for other classes.
The acid then equilibrates with bicarbonates in the blood, 'buffering' (minimising) the large changes in pH that would otherwise occur as dissolved CO2 and other acids entered and left the circulation.
See the Henderson-Hasselbach equation: http://www.chemistry.wustl.edu/~courses/genchem/Tutorials/Buffers/carbonic.htm

Drinking carbonated water will have very little effect (unless you drink it in excess) on your pH!

John

caveman1917
2010-Jul-27, 03:26 AM
An important point: For a closed system in dynamic equilibrium like CO2 gas dissolved in water, shaking would not affect the equilibrium. If you accept Henry's Law. The impression that the pressure is greater after shaking is very powerful, but misleading.

Here's an experiment that would disagree.

Take a bottle of coke. Open the bottle. Empty it so that only about 2/3 of the coke remains. Deform the bottle ('push' the sides inwards), decreasing its volume. Close the bottle. It will remain deformed (sides will remain pushed in). Let it sit overnight, so it can reach equilibrium. Note that it will still be deformed. Shake it. The sides will 'pop out', back to its original form/volume.

At least that's what it did with me.

The pressure increase is most definitely real.

And I can't get rid of the idea that we cannot apply Henry's law when we apply external forces to a system.

caveman1917
2010-Jul-27, 03:40 AM
Replace the bottle cap with a rubber balloon. When you shake the
bottle, the gas expands into the balloon, while the volume of the
liquid hardly reduces at all. With the bottle cap on, the gas can't
expand, so the pressure inside the bottle rises instead.

This also seems like a good experiment, and you won't need to wait for a whole night for equilibrium. I don't have any balloons in the house though, perhaps someone else might test this and report back? :)

BioSci
2010-Jul-27, 06:31 PM
Ta-daa! (http://www.wonderquest.com/spewing-soda-can.htm)

A pretty good explanation here! Lots of contributors above have a good piece of the answer, and this pulls it all together. Especially the influence of bubbles.

An important point: For a closed system in dynamic equilibrium like CO2 gas dissolved in water, shaking would not affect the equilibrium. If you accept Henry's Law. The impression that the pressure is greater after shaking is very powerful, but misleading.

The problem with citing Henry's Law is that it applies to an equilibrium situation - shaking of a soda can results in CO2 gas coming out of solution and raising the pressure - the can is no longer at equilibrium.

But the increased outflow when the can is opened is not due (directly) to the increased pressure - but rather from the presence of micro CO2 bubbles - which can then allow additional CO2 to escape and expand explosively when pressure is released

This cite has a nice discussion of a class-based experiment to actually measure the increased pressure in a soda can with stress sensors (5 sec of shaking increased pressure 10-20%). (http://docs.google.com/viewer?a=v&q=cache:AsnQ_C2qOCQJ:technologyinterface.nmsu.edu/Spring06/12_Dues-Accepted/index.pdf+shaken+can+pressure&hl=en&gl=us&pid=bl&srcid=ADGEESg0lOae_9ReIo8DOzhmieq2rUQ3IckdV2RSK-bNLscR1rW2Ss7dS5IWIveqM13t0_ILtMQLw5fGKxSZUZ2fxOMF h3uQ2HT_HYQHWJptuOLOSob8jX751Gyezf0MGS3UrpuIQQGa&sig=AHIEtbRE6KM4N6O7kfWEz4GKLUzPbYJjWw)

caveman1917
2010-Jul-27, 06:40 PM
Then the remaining question is, how exactly does shaking cause CO2 to move out of the liquid?
I had a wild guess in post #6 (http://www.bautforum.com/showthread.php/106045-Bottle-Pressure.?p=1764058#post1764058) but wether that's correct i do not know.

mugaliens
2010-Jul-27, 06:46 PM
CO2 in water definitely dissolves to form carbonic acid - it's a significant part of the flavor of many sodas.

Correct: "For a CO2 pressure typical of the one in soda drink bottles ( ~ 2.5 atm), we get a relatively acid medium (pH = 3.7) with a high concentration of dissolved CO2. These features contribute to the sour and sparkling taste of these drinks." - Source (http://en.wikipedia.org/wiki/Carbonic_acid#pH_and_composition_of_a_carbonic_aci d_solutions)

Drinking carbonated water will have very little effect (unless you drink it in excess) on your pH!

Our bodyies are wonderfully adapted towards keeping blood pH within appropriate limits: about 7.35 to 7.45. Increased acidity can be caused by many things, ranging from lactic acide buildup to ketosis and others, is quickly balanced in the healthy human by increased exhalation of carbon dioxide from the lungs. Interestingly, blood is heavily buffered against pH shifts.

Those of us who frequent high altitudes such as climbing 14ers are aware of the reverse problem, where we blow off CO2 in such volumes in our desire for oxygen that we frequently get into a situation of respiratory alkalosis, which results in bicarbonate excretion from the kidneys, one of the reasons why proper hydradtion is so critical in high-altitude efforts.

mike alexander
2010-Jul-27, 08:21 PM
Here's an experiment that would disagree.

Take a bottle of coke. Open the bottle. Empty it so that only about 2/3 of the coke remains. Deform the bottle ('push' the sides inwards), decreasing its volume. Close the bottle. It will remain deformed (sides will remain pushed in). Let it sit overnight, so it can reach equilibrium. Note that it will still be deformed. Shake it. The sides will 'pop out', back to its original form/volume.

At least that's what it did with me.

The pressure increase is most definitely real.

And I can't get rid of the idea that we cannot apply Henry's law when we apply external forces to a system.

OK, I think I see this. Yes, the pressure increase is real, but the rules of the game have changed.

When you open the bottle you equalize the pressure with the atmosphere. You also let in some ordinary air (which has a much lower water solubility). Now you squeeze the bottle and cap it. It contains mostly ordinary air at 1 atm and a solution of CO2 that is not yet in equilibrium. The concentration of CO2 in water and in the space over the water will now equilibrate. Remember that the partial pressure of CO2 in the gaseous phase is independent of the pressure of other gases present. When equilibrium is reestablished the pressure in the bottle will be the sum of the atmospheric gases plus the partial pressure of equilibrated CO2. Since this is greater than 1 atm the pressure inside will pop out the bottle sides.

So, the pressure increases, but it does not violate Henry's law.

Anyhow, that feels right to me.

Jeff Root
2010-Jul-27, 09:39 PM
OK, I think I see this. Yes, the pressure increase is real, but the
rules of the game have changed.

When you open the bottle you equalize the pressure with the
atmosphere. You also let in some ordinary air (which has a much
lower water solubility).
There is no need for any significant amount of air to enter the
bottle. Done carefully, the amount of air entering should be
much less than 1% of the bubble, and way less than 1% of the
carbon dioxide still in solution.

Now you squeeze the bottle and cap it.
I plan to squeeze the bottle in a vertical position, allowing some
liquid to overflow, and put the cap on without letting any air in.

It contains mostly ordinary air at 1 atm and a solution of CO2 that
is not yet in equilibrium. The concentration of CO2 in water and in
the space over the water will now equilibrate. Remember that the
partial pressure of CO2 in the gaseous phase is independent of
the pressure of other gases present.
When in equilibrium.

When equilibrium is reestablished the pressure in the bottle will be
the sum of the atmospheric gases plus the partial pressure of
equilibrated CO2. Since this is greater than 1 atm the pressure
inside will pop out the bottle sides.

So, the pressure increases, but it does not violate Henry's law.
I'd say the reason it doesn't violate Henry's law is simply that
Henry's law applies to stuff in equilibrium, and adding air has
nothing to do with it.

-- Jeff, in Minneapolis

caveman1917
2010-Jul-27, 11:07 PM
When equilibrium is reestablished the pressure in the bottle will be the sum of the atmospheric gases plus the partial pressure of equilibrated CO2. Since this is greater than 1 atm the pressure inside will pop out the bottle sides.

No, you may have misread somewhere, it doesn't pop out when it regains equilibrium. The sides stay 'popped in' at all times, until you shake it. I left it overnight to give it time to go to equilibrium, but it was still deformed then. Only when shaking it do the sides pop out again. The act of shaking is what increases pressure. Which was what had to be shown. It has nothing to do with the air.

ETA: i'm really getting convinced you're trying to apply a law where it can't be applied, ie to an open system.

caveman1917
2010-Jul-27, 11:16 PM
I plan to squeeze the bottle in a vertical position, allowing some
liquid to overflow, and put the cap on without letting any air in.

Take care to squeeze it enough, almost 'side to side', because when going to equilibrium it may rise pressure a little, so you need to make sure that small effect doesn't pop everything out again before the experiment even started.

ETA: i'm not sure it will work if you only leave liquid in there, but let's see... how much fun the scientific method can be, now all we need is a bookmaker :)

mike alexander
2010-Jul-28, 08:58 PM
No, you may have misread somewhere, it doesn't pop out when it regains equilibrium. The sides stay 'popped in' at all times, until you shake it. I left it overnight to give it time to go to equilibrium, but it was still deformed then. Only when shaking it do the sides pop out again. The act of shaking is what increases pressure. Which was what had to be shown. It has nothing to do with the air.

ETA: i'm really getting convinced you're trying to apply a law where it can't be applied, ie to an open system.

No, I'm just trying to ferret out all the conditions in an experiment.

I think I'll go have a Pepsi.

caveman1917
2010-Jul-28, 11:08 PM
No, I'm just trying to ferret out all the conditions in an experiment.

I think I'll go have a Pepsi.

The effect you quote is certainly real. The partial pressure of the CO2 will increase as it goes back to equilibrium, so the volume of the gas will increase by a factor <2. However if we take care to have squeezed the bottle enough that we still have plenty of 'spare room' after this expansion, you'll see that by shaking it it'll pop out all the way again.

caveman1917
2010-Jul-28, 11:18 PM
Here's a quick version of the experiment, where you won't need to wait for it to regain equilibrium, and it'll take away the air-argument.
I don't have any bottles of soda left now, will buy some tomorrow if someone else doesn't beat me to the experiment.

Take a bottle of soda.
Hold it upside down. (preferably over the sink :) )
Start squeezing it.
Slightly open the cap, so the soda can start to get out. It will splash a bit i suppose.
It's important to keep squeezing it, so it won't 'burp' any air upwards. Don't open the cap too much, just enough so the liquid can get out a bit.
After a while, when squeezed enough, turn the cap back to fully closed.

Now, according to Mike's prediction, we are still in equilibrium. The volume, pressure and composition of the gas is still the same.
All we did was remove some volume of liquid. But the volume of the liquid is irrelevant to Henry's law.

So we can immediately shake it.
If it pops out, you're wrong. If it doesn't, i'm wrong :)

caveman1917
2010-Jul-28, 11:43 PM
As was to be expected, curiosity gotten the better of me, and i drove to the night shop to get some bottles of coke.

I did the experiment in the post above, and found that the sides indeed popped back out. The pressure does increase by shaking.

I think the best explanation is that the act of shaking means applying external forces to a system, so that it cannot be considered closed anymore. Hence Henry's law isn't valid from the moment it gets shaken.

Ken G
2010-Jul-29, 04:00 AM
I hadn't even heard of Henry's law prior to reading this thread, but I found the linked answer pretty convincing, and it did invoke Henry's law. The key issue at this point seems to be the question of whether or not the pressure increases when you shake soda. The link said it doesn't, and I think they are right. So why does the soda zoom out when shaken, and why does the squeezed bottle puff out?

I think the soda zooms out for the same reason that you will overflow your glass if you pour the soda too quickly. When you first open a can of soda, the pressure immediately drops from 2 atm to 1 atm, that much we can all agree. This means Henry's law, which is an equilibrium condition, is immediately unsatisfied-- the dissolved CO2 is unhappy, half of it wants to come out of solution in order to reach a new equilibrium. This "unhappiness" is why soda is fizzy, and once it gets down to half its earlier concentration, the soda is completely "flat." But it takes time to come out of solution, unless something helps it along-- apparently, one thing that helps it along, and drastically reduces the time it takes to come out of solution, is the presence of vortexes from shaking or pouring, coupled with the presence of nucleation centers. It is the rapid escape of over-dissolved CO2 that causes the explosion out of the bottle, or the overflowing the glass, not a pressure increase-- the pressure is still 2 atm before you open it, and 1 atm right after, the difference is in how the soda responds to that sudden change-- how long it takes to "go flat."

So what about if you have already opened the plastic bottle, and squeezed it? This is going to be a somewhat different situation, because the soda starts out at 1 atm, not 2 atm, after you reclose the lid. You have 2 atm worth of CO2 dissolved (minus whatever came out of solution the first time you opened it, but perhaps that's not very much), but it's only under 1 atm of pressure, so it's going to slowly come out of solution (or faster if you shake it). Either way, it will attempt to restore 2 atm, and will succeed if the empty space in the bottle is pretty small (if there's a lot of empty space, on the way to returning to 2 atm most of the CO2 will come out of solution, so it won't ever get back to 2 atm). So whether you shake it or not, it should unsqueeze the bottle as it tries to return to 2 atm-- shaking it just makes it happen faster. So my guess is, in your experiment, you simply didn't wait long enough before you shook it-- you sped up a process that was on its way to happening anyway.

caveman1917
2010-Jul-29, 04:14 AM
So my guess is, in your experiment, you simply didn't wait long enough before you shook it-- you sped up a process that was on its way to happening anyway.

So your argument is that the shaking didn't start of from an equilibrium position? That would indeed invalidate the experiment. I did wait 8+ hours for it though, that would seem enough. [ETA: given that the same amount of time is enough for a glass of soda to lose all it's carbon to the air]

The main argument here seems to be over wether, starting from an equilibrium, does the shaking ruin that equilibrium or not.

But then how about the second experiment? The CO2 remained at all times at the same 2bar pressure, and the composition didn't change.

To be sure, i'll do the first experiment again, leaving only a very small volume of gas in the bottle, so leaving overnight should be enough for the CO2 to come out of solution back to 2bar.

publius
2010-Jul-29, 04:52 AM
But then how about the second experiment? The CO2 remained at all times at the same 2bar pressure, and the composition didn't change.

And you have the pressure readings to prove it? :) You can't eyeball and feel pressure that way. The only way to be sure is to acurately measure it, and if you set that up, you don't need to bother with squeezing it in, just shake it and watch the pressure reading.

Here's what I just did. I opened a bottle of Pepsi, sitting in the refrigerator. Phsssst! I just cracked the cap to allow the excess pressure to escape and then closed it back it tight. Very little air should have exchanged. Squeezing it verified the pressure was lower. That is, a 2atm bottle is harder to squeeze (but can still be squeezed) that a 1atm bottle.

Shaking it causes it quickly to jump back up to 2 atm, as verified by the "harder to squeeze" test. Harder to squeeze only tells me the pressure increased, not by how much, although I would guess it went back to baseline.

Now, I take another bottle of Pepsi and do not open it. I shake and shake and shake. A head of foam develops, but the bottle does not feel appreciably harder to squeeze. I'll grant you a transient increase in pressure due to dynamic, out-of-equilibrium conditions during the violent shaking. But the equilibrium pressure doesn't seem to increase. The head of foam is still there even after I typed all this, so I expect it will take some time for the "bubble tendency" to subside before it would be safe to open. :)

Second, we're not considering thermal transfers. I don't know how significant that would be. Holding a cold bottle in a hot hand will transfer some heat, but I have no idea if that would be significant.

I'll tell you what will increase the pressure, though. Freezing it. :) I ruptured a can once by doing that. I put it in the freezer to cool it quickly and forgot about it. It froze solid and the can was bulging. I took it out and put it in the sink and ran a little hot water over it. Pssssht! it went, rupturing the weak spot around the tab opening making a small leak. That leak spewed foam all over the sink and my arms. When it was over, I was left with a frothy chunk of slush ice in the can.

-Richard

caveman1917
2010-Jul-29, 05:16 AM
And you have the pressure readings to prove it? :)

No i don't :).
However the given explanation would predict it doesn't. The gas always remains in the same local state with regards to the liquid. One can eye-ball the volume remains the same. All that happens is removing some liquid out of the 'bottom'.

The only way to be sure is to acurately measure it, and if you set that up, you don't need to bother with squeezing it in, just shake it and watch the pressure reading.

Now, I take another bottle of Pepsi and do not open it. I shake and shake and shake. A head of foam develops, but the bottle does not feel appreciably harder to squeeze.

True, however i'm also not suggesting a pressure increase by any large factor.
What i'm suggesting is a slight increase in equilibrium pressure due to shaking giving a higher average velocity to the constituent molecules of the liquid, which would be a thermal effect in the end. Since we obviously can't increase temperature of a liquid very much by shaking it, the increase in pressure would also not be by any large amount. But it would be there nonetheless. That's what i think the readings of 10-20% increase in the stress-sensor experiments given above would come from.

Second, we're not considering thermal transfers. I don't know how significant that would be. Holding a cold bottle in a hot hand will transfer some heat, but I have no idea if that would be significant.

Right. I suppose these would be neglegible since we're talking about mere seconds of contact, but i guess one can take the bottle by the cap and shake it that way to even minimize those effects.

caveman1917
2010-Jul-29, 05:30 AM
Ok, i set up another bottle, leaving only a volume of gas of about three 'caps' in the bottle.

How much time does it need to sit so people won't argue that it wasn't given enough time to go to equilibrium? :)

caveman1917
2010-Jul-29, 05:34 AM
If publius is correct, then the transient increase in pressure would make the sides pop out again, but the pressure decrease that happens afterwards, with the volume remaining the same, would suggest the gas part would end up at <1atm. Opening it would then suggest an inwards flow of air, which could be tested by enclosing it with a balloon before taking of the cap.

ETA: nevermind, got this one totally wrong obviously, i guess it's time to call it a day and finally head to bed, shouldn't be trying to think about this kind of stuff till 7am :)

publius
2010-Jul-29, 06:18 AM
This just reminded me of something. Acetylene storage. While electric welding has long since displaced it, one still needs it for the cutting torch and perhaps some gas welding just for fun. I've got a rig. Now, O2 is stored as a gas under high pressure in a thick walled cylinder. One cannot store acetylene that way as it unstable under pressures over about 30psi or so, undergoing some exothermic reaction which can blow things up good. You don't store much gas at 30psi.

So what they figured out, and this was around 100 years ago, I think, was to dissolve it in acetone. Now, 30psi dissolved therein still doesn't get you enough, but if you use a porous material with very tiny holes throughout, the explosion risk is eliminated (something about if the capillaries are small enough, the reaction can't happen). Now, they still had problems with the material breaking down and the holes getting too big, but they soon fixed that.

So, an acetylene cylinder is filled with a sturdy porous material, wetted with acetone which then dissolves about 145 cubic feet in the typical size.

-Richard

mugaliens
2010-Jul-29, 06:32 AM
Very interesting tidbit, Richard - I did not know that!

My son and I picked up some "hydro-crackers" earlier this summer. They're waterproof firecrackers, and work great when thrown into a bucket of water.

Well, I had a bottle of Diet Pepsi sitting in the fridge, and since I rarely drink the stuff, I decided to conduct a little "experiment" of my own, namely dropping the cracker into the bottle while trying to get the top back on it time.

I was able to get the top on before the cracker blew, but the bottle ruptured anyway, and with that sort of concusion, the liquid came gushing out of the crack and sprayed all over the bathroom.

Not much of an experiment, mind you. No hypothesis, no controls... But it was fun! Not the cleaning up part, though.

Ken G
2010-Jul-29, 07:46 AM
It sounds like we all now agree that the reason the soda shoots out so dramatically is because half the CO2 is trying to come out of solution whenever you first open a can, but if it is shaken, it happens much faster. Since Henry's law tells us why half the CO2 in a previously 2 atm can wants to come out of solution, Henry's law provides the underlying reason for the explosion. But it is more like a signpost toward the state the system is trying to achieve, rather than a moment-by-moment description of what state the system is in. Thus, we need to separate the ultimate equilibrium from the timescale to reach it-- both are equally important considerations. So I couldn't claim Henry's law requires that there's no pressure increase at all when you shake soda, because Henry's law is an equilibrium law, so takes time to take effect. Short-term pressure increases might occur for reasons that have nothing to do with Henry's law. The external link claimed there was no pressure increase, but they might just mean there was no increase significant enough to be important for a dramatic soda explosion, so just what happens when you shake a closed can remains a somewhat open question.

Another interesting point that came up was, after you open a bottle, and it drops to 1 atm, and then you close it off again, how long does it take to return to 2 atm? And then what happens when you shake it? When you squeeze a bottle at 1 atm, and it slowly returns to 2 atm, it has to completely re-expand to support the 2 atm. That tells you how much CO2 must come out of solution-- enough to both re-expand the bottle, and get the partial pressure up to 2 atm. So it seems it must completely re-expand to get to a new equilibrium-- if it hasn't completely re-expanded, it hasn't yet reached it's Henry's law equilibrium-- assuming the empty volume is small.

But there's another possibility-- the CO2 craps out before the volume re-expands, and the whole thing is still at 1 atm. Then if you shake it, it's like an ocean wave breaking-- you get some CO2 coming out of solution even at 1 atm. If the bottle isn't fully puffed, it's at 1 atm, we just don't know if it is in Henry equilibrium yet or not. If you know you haven't lost much CO2 in solution, and you know the volume of the empty part of the bottle is small, and it isn't fully puffed, then you know it is at 1 atm and it is not yet in Henry equilibrium-- you have to wait longer.

Either way, the experiments show that the bottle puffs out when shaken, which means that both the volume and the amount of CO2 gas increases upon shaking. Note that does not require a pressure increase-- that could all happen at 1 atm. Only if the bottle fully repuffs out should you expect the pressure to exceed 1 atm. But the need for more gas to come out of solution is there in any case, whether the pressure changes or not. So it doesn't sound like pressure increase is ever the explanatory issue, though some of it may occur as an after-effect. The explanatory issue has to be why CO2 comes out of solution when you shake it, and the answer has to take into account whether or not you were already in Henry equilibrium before you shook it.

caveman1917
2010-Jul-30, 12:07 AM
It sounds like we all now agree that the reason the soda shoots out so dramatically is because half the CO2 is trying to come out of solution whenever you first open a can, but if it is shaken, it happens much faster. Since Henry's law tells us why half the CO2 in a previously 2 atm can wants to come out of solution, Henry's law provides the underlying reason for the explosion. But it is more like a signpost toward the state the system is trying to achieve, rather than a moment-by-moment description of what state the system is in.

Yes. The pressure increase i was trying to show wasn't meant to be the causal effect for the squirting of the soda bottle, it is too small for that. The bubbles explain that effect.
The claim i took issue with was that shaking a bottle does not open the system, and Henry's law could be applied throughout the entire process.

When you squeeze a bottle at 1 atm, and it slowly returns to 2 atm, it has to completely re-expand to support the 2 atm.

But there's another possibility-- the CO2 craps out before the volume re-expands, and the whole thing is still at 1 atm. Then if you shake it, it's like an ocean wave breaking-- you get some CO2 coming out of solution even at 1 atm.

Either way, the experiments show that the bottle puffs out when shaken, which means that both the volume and the amount of CO2 gas increases upon shaking. Note that does not require a pressure increase-- that could all happen at 1 atm. Only if the bottle fully repuffs out should you expect the pressure to exceed 1 atm.

This claim (when shaking it needs to expand completely before pressure will go up over 1atm) seems to be falsified by this experiment:

Open the bottle, it's now 1atm, squeeze it and close it. Immediately shake it, while tightly holding both hands around the bottle. You will feel pressure goes up (by how much is uncertain, but at least enough that i couldn't readily hold it back) before it expanded fully.

I just tried this one again, this time putting a lot of force on the bottle. It seems there is a maximum pressure it will reach, i wasn't able to withstand it expanding a little, but was able to withstand it expanding fully. The final pressure seemed on the same scale as the 2atm normally found. Wether it's exactly 2atm, or a little more, one can't tell of course. [ETA: the final pressure thing may be false, i was using the same bottle for a couple of times, and found a lot of the CO2 had already gone out of the liquid anyway. So that final pressure may have just been the result of that, not any maximum inherent to the process]

The claim that it will need to expand fully before reaching Henry-equilibrium by itself also seems a little shaky. Suppose we tape the squeezed bottle of tightly. That would mean it would never be able to reach equilibrium again.

I still had an experiment running, so for what it's worth, i thought i might as well just report it.
Setup: squeezed at 1atm, volume of gas ~3 caps. Left to sit for 20 hours. The gas had expanded by a factor of about 2-3. Shook it, puffed out completely at a pressure which seemed around 2atm, or somewhat more. I wish i had thought of trying to squeeze it even more before shaking, to see at what pressure the bottle was at that point.

Ken G
2010-Jul-30, 03:16 AM
The claim i took issue with was that shaking a bottle does not open the system, and Henry's law could be applied throughout the entire process.Yes, Henry's law only applies in detail after you wait awhile. Moment-by-moment, it won't hold, it will just indicate the direction things are going.

This claim (when shaking it needs to expand completely before pressure will go up over 1atm) seems to be falsified by this experiment:

Open the bottle, it's now 1atm, squeeze it and close it. Immediately shake it, while tightly holding both hands around the bottle. You will feel pressure goes up (by how much is uncertain, but at least enough that i couldn't readily hold it back) before it expanded fully.That's a different situation, because you are holding it. It is the fact that you are squeezing it that allows the pressure to exceed 1 atm. I was talking about an unheld bottle, which is always at 1 atm unless it is fully puffed out (at which point the bottle can "be squeezing" itself). What I'm saying is that if you pop open the bottle, let a little soda out to create an empty bubble, squeeze to minimize that empty bubble, and recap the bottle, then you have a bottle at 1 atm. If you then shake it, or just wait for it, either way it will still be a 1 atm bottle until it repuffs fully. In particular, if you have not waited long enough for it to repuff fully, it is not in Henry equilibrium yet, as it is still at 1 atm but needs to be nearly 2 atm to be in equilibrium.

I just tried this one again, this time putting a lot of force on the bottle. It seems there is a maximum pressure it will reach, i wasn't able to withstand it expanding a little, but was able to withstand it expanding fully. The final pressure seemed on the same scale as the 2atm normally found. Yes, I'd say that's a description of using shaking to shorten the time it takes to re-establish the Henry equilibrium that it would have eventually reached anyway, but how long it would have taken I don't know. 8 hours seems like it should be enough (or the 20 hours in your next experiment), but apparently not-- not if the soda is still not flat, and the bottle hasn't fully repuffed (like if it still has visible dimples in it).
The claim that it will need to expand fully before reaching Henry-equilibrium by itself also seems a little shaky. Suppose we tape the squeezed bottle of tightly. That would mean it would never be able to reach equilibrium again.To clarify, the statement that it must fully repuff to reach equilibrium holds for an unsqueezed bottle. Taping it is like redefining the meaning of "fully puffed." The volume available to the empty part of the bottle is a bit of a red herring-- if we have the same dissolved CO2 content as before (the soda is still "fizzy"), then it will always reach Henry equilibrium when the partial pressure of CO2 in the bubble is 2 atm. That is true independent of the volume that the empty bubble must reach to achieve 2 atm-- unless that volume is so large it sucks out a significant amount of the dissolved CO2 on the way to Henry equilibrium (a possibility we eliminate by insuring the soda is not flat, and keeping the bubble volume a small fraction of the bottle).

Setup: squeezed at 1atm, volume of gas ~3 caps. Left to sit for 20 hours. The gas had expanded by a factor of about 2-3.But if the bottle is not fully repuffed by this point, the pressure must still be 1 atm, because a collapsed bottle cannot support anything higher. So it hadn't reached Henry equilibrium, even after 20 hours-- surprisingly! This must relate to the reason we are advised to squeeze our bottles to keep the soda fizzy-- it extends the equilibration time, though this is more of an issue for bottles that have large empty spaces so we couldn't get them to 2 atm without the soda going flat. When the Henry equilibrium will make our soda flat, the best we can do is lengthen the time it takes to reach it.
Shook it, puffed out completely at a pressure which seemed around 2atm, or somewhat more.That must be the first reappearence of the Henry equilibrium, modulo any overshooting that would have to readjust with time.

caveman1917
2010-Jul-30, 03:21 AM
I had misunderstood Ken. I thought the claim you made (with saying it needed to expand fully) was re the volume of the gas, not the counterpressure of the bottle. You are obviously correct in this sense.

I'm also very suprised even 20 hours isn't enough.

This brings up an interesting question, do we in real life ever deal with cans/bottles in Henry-equilibrium? There seems to be so much time needed for it to reach it. The volume after 20 hours still had not even gotten to half of what it should've been, if it even gotten to 1/4. And even then it would still needed to start all over again from 1atm to 2atm...

ETA: that is assuming disturbing the bottle does indeed ruin equilibrium, albeit only on a small scale, but it just seems to need so much time to recover

caveman1917
2010-Jul-30, 03:32 AM
In the end, it seems BioSci is the winner here (http://www.bautforum.com/showthread.php/106045-Bottle-Pressure.?p=1768517#post1768517) :)

Ken G
2010-Jul-30, 03:34 AM
I'm also very suprised even 20 hours isn't enough.Which raises the next question-- how long does it take? It might seem like it shouldn't take as long as an uncapped bottle takes to go completely flat, because if it is uncapped, half the CO2 has to leave the solution to reach equilibrium at 1 atm, whereas if it is capped and the bubble is small, you only need a small fraction of that to leave solution to get 2 atm. But maybe the fact that capping it reduces air currents that might help it leave solution, or maybe particles from the air get into an uncapped bottle and greatly speed up the equilibration process?

I don't know, because even when I cap a plastic bottle, it seems to go flat in a few days. But maybe that's because I get thirsty, and don't just leave it be for those few days! It might be that it takes very large steps toward Henry equilibrium every time you uncap it, whereas if you leave it alone, and the empty volume is small, it takes a very long time to equilibrate. I see another experiment coming!

Ken G
2010-Jul-30, 03:39 AM
In the end, it seems BioSci is the winner here (http://www.bautforum.com/showthread.php/106045-Bottle-Pressure.?p=1768517#post1768517) Not to quibble, but I think the statement

The problem with citing Henry's Law is that it applies to an equilibrium situation - shaking of a soda can results in CO2 gas coming out of solution and raising the pressure - the can is no longer at equilibrium.
is pretty misleading. If the can is at 2 atm and unopened, shaking it will not really take it out of Henry equilibrium (maybe 10% or so, as in that link, but it wouldn't make any real difference if it was 0%-- we've agreed that neither the pressure increase, nor the concomitant disequilibrium, are essential features of the soda explosion you'll get). The crucial impact of the shaking is to set the stage for a much more rapid return to equilibrium, at 1 atm, when the can is then opened. So ironically, it is not that shaking it gets you away from equilibrium that is the key here, it is that shaking it greatly speeds up equilibration when you open it-- the important disequilibrium is the 2 atm vs. 1 atm, not the 2 atm vs. whatever it is after you shake it.

I think that external link was right when it said that it is hard to find the correct explanation until you have visited that site!

caveman1917
2010-Jul-30, 03:41 AM
There is something here that might delay things though. Since we let regular air in, we are talking about the increase of the partial pressure of CO2. Now as i understand, in regular air CO2 is only a small fraction. That would seem to suggest that the increase of the partial pressure of that small fraction from 1atm to 2atm, would not change the total pressure of the gas by very much. So the total pressure of the gas might not have been enough to expand the bottle fully, even though the partial pressure of the CO2 was already at 2atm. It may in all even have been <1.1atm.
We may have been in Henry-equilibrium all along.

caveman1917
2010-Jul-30, 03:46 AM
Not to quibble, but I think the statement

is pretty misleading. If the can is at 2 atm and unopened, shaking it will not really take it out of Henry equilibrium (maybe 10% or so, as in that link, but it wouldn't make any real difference if it was 0%-- we've agreed that neither the pressure increase, nor the concomitant disequilibrium, are essential features of the soda explosion you'll get). The crucial impact of the shaking is to set the stage for a much more rapid return to equilibrium, at 1 atm, when the can is then opened. So ironically, it is not that shaking it gets you away from equilibrium that is the key here, it is that shaking it greatly speeds up equilibration when you open it.

I think that external link was right when it said that it is hard to find the correct explanation until you have visited that site!

But that's also not what he's saying, if you read on. He's not claiming the pressure increase is a feature of the soda explosion. I do think the statement he makes is in essence correct, shaking will increase pressure (albeit small and irrelevant - but it increases nonetheless). Wether that takes it out of equilibrium depends on the pressure in the liquid, so arguably that part of the statement can be wrong -- if the liquid has an equal increase in pressure, equilibrium is not broken.

I meant that he's the winner in the sense that he was the first one to state basically the whole story: pressure does increase upon shaking, but it's the bubbles that create the squirting effect.

Ken G
2010-Jul-30, 03:51 AM
There is something here that might delay things though. Since we let regular air in, we are talking about the increase of the partial pressure of CO2.Yes, that's a nasty complication, but just means that the pressure actually goes above 2 atm before you recover equilibrium.

Now as i understand, in regular air CO2 is only a small fraction. That would seem to suggest that the increase of the partial pressure of that small fraction from 1atm to 2atm, would not change the total pressure of the gas by very much.Actually, it would mean the pressure could go as high as 3 atm, but it depends on how small you make the empty bubble. The amount of gas here is essentially the pressure times the volume, so if you had regular air at 1 atm in the empty bubble of initial volume Vo, Henry equilibrium could not return until 2 atms of CO2 had become gaseous. But that will require returning the bottle to fully repuffed status, which might require a volume Vf. So the amount of gas would end up 1 atm * Vo of regular air, plus 2 atm * Vf of CO2. The pressure requires dividing by Vf, so the final pressure is 1 atm * (Vo/Vf + 2). That's why volume is something of a red herring here, all you need is that the final partial pressure of CO2 will be 2 atm, and that must fully repuff the bottle to support that.

publius
2010-Jul-30, 03:54 AM
I wonder how much temperature has to do with it. Will a room temperature bottle come back to equilibrium much faster than a refrigerator cold bottle? I left a 16 oz bottle in the refigerator since this morning which I had taken a swig out of, and it was still only slightly pressurized this evening.

And in playing around, I discovered that shaking it to build the pressure back up will quickly flatten it, when the liquid level is down enough that you have significant air space.

And I think we all can easily see why big 2 liter bottles will go flat faster than smaller bottles -- much more air space as you use it up. Just had an idea -- make bottle with a little piston or other volume reducing method to keep air space minimized. I'm sure that would cost much more than it was worth, otherwise they'd probably do it already.

Or hey, go to your local welding store, get a bottle of CO2, and rig a little manifold and keep your soda pressurized at 2atm all the time. :lol:

I'm mostly joking, but that wouldn't be too hard to rig, really. Get yourself a little pressure vessel, tap it on the bottom, and put a regulator to keep CO2 pressure above it at 2atm. Or even more if you want a really fizzy kick. Reckon there would be a market for such a contraption? :)

-Richard

mike alexander
2010-Jul-30, 03:57 AM
After opening the bottle and recapping it, realize that the pressure will not return to 2 atm at equilibrium.

caveman1917
2010-Jul-30, 03:58 AM
Yes i had kept the volume constant while quickly trying to figure out what it would mean to have air instead of CO2, i really shouldn't be doing this kind of thinking so late at night/morning :)

caveman1917
2010-Jul-30, 04:00 AM
After opening the bottle and recapping it, realize that the pressure will not return to 2 atm at equilibrium.

Why not?

caveman1917
2010-Jul-30, 04:01 AM
Reckon there would be a market for such a contraption? :)

There's a market for everything, demand is created, not inherent after all ;)

Ken G
2010-Jul-30, 04:01 AM
But that's also not what he's saying, if you read on. He's not claiming the pressure increase is a feature of the soda explosion. I do think the statement he makes is in essence correct, shaking will increase pressure (albeit small and irrelevant - but it increases nonetheless). Wether that takes it out of equilibrium depends on the pressure in the liquid, so arguably that part of the statement can be wrong -- if the liquid has an equal increase in pressure, equilibrium is not broken.
My point was that even mentioning the way shaking can take you out of Henry equilibrium is a red herring to the soda explosion-- because what really takes you out of Henry equilibrium is opening the can. Shaking mostly affects what happens after that big disequilibrium appears (and BioSci was right that the presence of tiny bubbles then matters, since the external link pointed to the importance of how vortices help support those tiny bubbles).

I meant that he's the winner in the sense that he was the first one to state basically the whole story: pressure does increase upon shaking, but it's the bubbles that create the squirting effect.Yes, and all the more impressive that he gave that argument before the appearance of the external link. His minor deflection about the disequilibrium elements of shaking, which are only relevant to the separate issue of small pressure increases, could easily have been in reference to that other discussion going on simultaneously, so was not intended to deflect from the importance of what happens to the Henry equilibrium when you open the can. So I retract any accusations of "misleading", that was more in my misinterpretation of his intent.

Ken G
2010-Jul-30, 04:04 AM
After opening the bottle and recapping it, realize that the pressure will not return to 2 atm at equilibrium.
There is some lost concentration of dissolved CO2, but that's not an important detail. We can imagine attempting to minimize such losses, and if it returns to 1.9 atm instead of 2 atm, it's close enough for government work.

caveman1917
2010-Jul-30, 04:05 AM
BioSci, you can pick up your prize at Richard's, the very first prototype :D

I'm mostly joking, but that wouldn't be too hard to rig, really. Get yourself a little pressure vessel, tap it on the bottom, and put a regulator to keep CO2 pressure above it at 2atm. Or even more if you want a really fizzy kick. Reckon there would be a market for such a contraption? :)

Ken G
2010-Jul-30, 04:15 AM
I wonder how much temperature has to do with it. Will a room temperature bottle come back to equilibrium much faster than a refrigerator cold bottle?(Don't we have enough control variables to play with already?) Just kidding, it's a good point.

I left a 16 oz bottle in the refigerator since this morning which I had taken a swig out of, and it was still only slightly pressurized this evening. I wonder if "backwash" might be important there (given that you said "swig"). My guess is, if you swig directly from the bottle, you better intend to drink it all-- because tiny particles from your mouth are going to end up in there, and rapidly increase the re-equilibration rate every time you open it again. The worst case scenario is to drink half the bottle right from the bottle, putting in lots of particulates, and then recap it. It will try to get back to 2 atm pretty quickly, perhaps, and that will sap a whole lot of that dissolved CO2 into that half empty bottle. That's when you discover how lousy most sodas taste when flat!

Just had an idea -- make bottle with a little piston or other volume reducing method to keep air space minimized. I'm sure that would cost much more than it was worth, otherwise they'd probably do it already. That would work, though the poor-man's version is squeezing a plastic bottle to remove as much empty space as possible. That won't work like yours, because it will still be at 1 atm, but it seems to greatly increase the time to Henry equilibration. That must be at least in part due to the fact that if you squeeze it down, you will at least have a partial CO2 pressure of 1 atm as it reexpands, whereas if you don't, the partial pressure of CO2 will take a long time to reach 1 atm. Thus it should roughly double the equilibration time to squeeze it down.

I'm mostly joking, but that wouldn't be too hard to rig, really. Get yourself a little pressure vessel, tap it on the bottom, and put a regulator to keep CO2 pressure above it at 2atm. Or even more if you want a really fizzy kick. Reckon there would be a market for such a contraption?Especially if 2 atm isn't the "magic fizziness"-- your contraption would let everyone set their own! I'm convinced that soda dispensers in fast-food joints have considerable variance from the 2 atm target, with sometimes unpleasant consequences.

caveman1917
2010-Jul-30, 04:18 AM
Yes, and all the more impressive that he gave that argument before the appearance of the external link.

No, it seems he gave it afterwards :(
Still impressive to put the whole thing together immediately though.

Ken G
2010-Jul-30, 04:25 AM
I'm most impressed by the external link itself. Rarely do I see explanations on the internet that cut through so much competing B.S. and get right to the key physical issues, even when they are quite subtle and advanced. He as good as said, "fasten your seat belts, because you won't find this anywhere else on the web." And that's hard to do in this day and age!

caveman1917
2010-Jul-30, 04:27 AM
Very true. Though i wished it would've stated a bit more exact what he meant by "i couldn't increase pressure" and continuously applying Henry's law. It did cause quite a debate here, but that is perhaps a good thing. Most fun too, squeezing and shaking bottles :)

publius
2010-Jul-30, 04:28 AM
I was sure others had thought of this. Apparently home carbonation is a big thing, and here's a DIY page:

http://www.truetex.com/carbonation.htm

Note his baling-wire and duct tape engineering, a man after my own heart. Note particular how he made fittings by drilling a hole in the regular plastic bottle caps.

There's apparently a "home soda fountain" market out there, with fancy (and expensive) premade equipment. So there is a market for this, just as I suspected.

-Richard

publius
2010-Jul-30, 04:33 AM
If you read that link, note temperature is very important as well as agitation (shaking!). Shaking helps the CO2 go *into* solution as well as it helps it come out, which makes sense. You want the water as cold as possible, which means 32F to get the most CO2 in there. I wonder what this means about the pressure of a room temperature bottle of carbonated water than was carbonated at 2atm and 32F?

-Richard

caveman1917
2010-Jul-30, 04:47 AM
Which raises the next question-- how long does it take?

When i get some new soda bottles, i'll set up an experiment :)

publius
2010-Jul-30, 04:47 AM
Well, shoot, in the \$100 - \$250 range for this retail soda fountain product:

http://www.sodastreamusa.com/default.aspx

Note the "consumer friendly" CO2 bottles -- I'm sure that's how they make their money over time. Me I'd prefer something with the big bottles from a welding supply of course, but that wouldn't do for the typical consumer.

-Richard

Ken G
2010-Jul-30, 04:59 AM
Very true. Though i wished it would've stated a bit more exact what he meant by "i couldn't increase pressure" and continuously applying Henry's law.Yes I'd forgotten about that, it does seem like he might have been flat out wrong there, one cannot assert equilibrium conditions categorically. Since it's not germane to the issue of the soda explosion, we'll cut him some slack.

Ken G
2010-Jul-30, 05:10 AM
I wonder what this means about the pressure of a room temperature bottle of carbonated water than was carbonated at 2atm and 32F? Henry's law comes to the rescue! Assuming it still applies at 32F (I couldn't guarantee that intermolecular forces near the phase transition don't mess it up), the partial pressure of CO2 cannot be affected by temperature. (It certainly is if you freeze the soda, as you mentioned, but that is sure to mess up Henry). However, the partial pressure of air will increase as it warms, so it should depend on how much air is in a standard can. Maybe not much, I don't really know.

Given the potential for intermolecular forces ruining Henry's law, I suppose it calls for yet another experiment! Anyone care to put a plastic bottle of soda in the Sun on a hot day and see if it is any harder to squeeze?

caveman1917
2010-Jul-30, 05:22 AM
Well Ken, i think i did my fair share of the experiments already :)
Doesn't need to be in the sun on a hot day, putting it on a radiator or suspending above the stove would work just as well i guess. (we dont get many sunny days over here, need to find other ways :))

And i still have another one coming to time how long it takes to regain Henry-equilibrium. Will need a control-bottle for that one too i guess :)

publius
2010-Jul-30, 05:22 AM
Henry's law comes to the rescue! Assuming it still applies at 32F (I couldn't guarantee that intermolecular forces near the phase transition don't mess it up), the partial pressure of CO2 cannot be affected by temperature.

Wait a minute -- the Henry coefficient must depend on temperature since the solubility varies with temperature, that of a gas in liquid usually decreasing with temperate. Thus the Henry "constant" must increase with temperature, meaning the partial pressure will increase with the temperature of the solution.

Wiki agrees on the Henry's Law page, assuming the authors know what they're talking about. :)

http://en.wikipedia.org/wiki/Henry%27s_Law#Temperature_dependence_of_the_Henry_ constant

-Richard

caveman1917
2010-Jul-30, 11:29 AM
One serious question still remains now. What caused the 10-20% pressure increase found in the classroom experiments?

Ken G
2010-Jul-30, 11:54 AM
Wait a minute -- the Henry coefficient must depend on temperature since the solubility varies with temperature, that of a gas in liquid usually decreasing with temperate. Thus the Henry "constant" must increase with temperature, meaning the partial pressure will increase with the temperature of the solution. You're right-- if we look at the value of "C" for CO2-- it's 2400 K, which means you have to get up to temperatures well above that to be able to neglect the effects of temperature changes. I thought when the external link, in its "Qualifications" section, talked about how high the solubility of CO2 is, it used that to not bother talking about the temperature that the soda was bottled at. But the solubility does not seem that high at all at 300 K, it seems like "high solubility" has to come at something more like 2400 K, if I'm understanding that Wiki formula.

Jeff Root
2010-Jul-30, 03:00 PM
One serious question still remains now. What caused the
10-20% pressure increase found in the classroom experiments?
Is that really in question? I may have missed something.
CO2 comes out of solution, becomes gas, the bottle prevents
the gas from occupying a larger volume than it had in the
liquid, so it compresses, increasing the pressure. No?

-- Jeff, in Minneapolis

Ken G
2010-Jul-30, 03:56 PM
The external link said the gas doesn't come out of solution when you shake it and don't open the can, all you get is vortices and tiny bubbles. But as has emerged from my discussion with publius on that Wiki entry, the external link also made some pretty questionable claims about the volume the CO2 would occupy at 1 atm if it all came out of solution (their answer says CO2 dissolves so well that it is basically still an ideal gas when in solution, and that's not corroborated by the "C" constant given in the Wiki).

publius
2010-Jul-30, 08:25 PM
You're right-- if we look at the value of "C" for CO2-- it's 2400 K, which means you have to get up to temperatures well above that to be able to neglect the effects of temperature changes. I thought when the external link, in its "Qualifications" section, talked about how high the solubility of CO2 is, it used that to not bother talking about the temperature that the soda was bottled at. But the solubility does not seem that high at all at 300 K, it seems like "high solubility" has to come at something more like 2400 K, if I'm understanding that Wiki formula.

That formula in Wiki is something of an approximation/ideal behavior thing as well (in the integration that derives it, I assume something complicated for real substances is assumed constant or something). Here's the first thing that I found Googling, a Word .doc file on CO2 in water with an empirical graph of the Henry coefficient vs temperature:

http://www.thuisexperimenteren.nl/infopages/Carbondioxide%20in%20water%20equilibrium.doc

Scroll down and you'll find a graph in figure 1. As seen there, the Henry coefficient usually goes through a maximum for gases in water, which appears to be somewhere aroun 140C for CO2 (I think the reactivity of CO2 in water to form carbonic acid further complicates things), and you can see that going from refigerator temperature to room temperature produces a significant change.

-Richard

Ken G
2010-Jul-30, 09:37 PM
The link didn't provide graphics on my Mac, but I was able to cull out the statement that 90 cc of CO2 [in gaseous equivalent volume at the partial pressure of the actual CO2] dissolves in 100 cc of H2O at room temperature. Because of Henry's law, we don't need to cite the pressure, that will always hold. That justifies the claim made in the previous external link that about 2 bottles worth of CO2 (at 1 atm) dissolves in 1 bottle at 2 atm. What's odd about this is that it is also the amount of CO2 that would be there if it were in the form of an ideal gas, rather than dissolved! So from the point of view of dissolving CO2 at room temperature (and I don't know how much that changes at refrigerator temps), the density of CO2 is the same in the fluid as in the empty space in the bottle, and the CO2 is "acting" pretty much like an ideal gas even when dissolved. What I didn't appreciate (because that original link suggested otherwise) is that this appears to be sheer coincidence for CO2 and room temperature, it's not something that would continue to hold at higher or lower temperatures (though I'm not sure how quickly it changes). The external link claimed this was because CO2 has such "high solubility", but that can't be right if what publius is saying (and what the Wiki said) is true-- it has to just be pure coincidence at room temperature, not a consquence of "high solubility", or else the temperature dependence can't be any big deal.

caveman1917
2010-Jul-30, 09:45 PM
Is that really in question? I may have missed something.
CO2 comes out of solution, becomes gas, the bottle prevents
the gas from occupying a larger volume than it had in the
liquid, so it compresses, increasing the pressure. No?

-- Jeff, in Minneapolis

This explanation seems to be contradicted by this: (at the time i didn't know, i forgot about that explanation, but it seems it's incorrect, so we're back at square 1 for explaining transient pressure increase)

What's odd about this is that it is also the amount of CO2 that would be there if it were in the form of an ideal gas, rather than dissolved! So from the point of view of dissolving CO2 at room temperature (and I don't know how much that changes at refrigerator temps), the density of CO2 is the same in the fluid as in the empty space in the bottle, and the CO2 is "acting" pretty much like an ideal gas even when dissolved.

publius
2010-Jul-30, 10:04 PM
The link didn't provide graphics on my Mac,

13493

-Richard

Ken G
2010-Jul-30, 10:18 PM
Yes, that's quite significant-- a 10% change in T (in K) can double the constant, and water before it boils can increase the constant by a factor of 5. What I can't tell is if high T means the CO2 has "low solubility", whereas near room temperature it achieves "high solubility" (and acts in effect like an ideal gas even when dissolved, in terms of its density when in equilibrium with a partial pressure). It seems like the chemisty of the dissolved CO2 is so complex it must be sheer coincidence that is has this behavior near room temperature, so it's not a consequence of "high solubility", but that external link certainly did seem to think this ideal-gas-like behavior was to be expected at high solubility.

caveman1917
2010-Jul-30, 10:20 PM
So any ideas on the cause of the transient pressure increase?

Ken G
2010-Jul-30, 10:33 PM
If it's there, it seems to be perhaps a 10% effect. At that level of precision, concepts like "high solubility" and "effectively ideal gas-like behavior" don't have much to say. I wouldn't be surprised if, at that level of precision, the extent to which the solubility isn't infinite, and the dissolved gas is not behaving ideally, means that some of it could come out of solution (as has been suggested above) just because of all the free energy provided by shaking. Equilibrium means that the entropy is maximized given the other constraints present, and shaking it could create a transient set of constraints that allow a higher entropy by having more CO2 in gaseous form. Then in time, that added free energy dissipates into work and ultimately heat, the heat leaves the system to reach temperature equilibrium, and the maximum entropy is back to its previous lower level (because of the heat transport out) and represents less CO2 in gaseous form, returning to its previous pressure.